Question

A rocket is fired vertically upwards with a net acceleration of 4 $m/s^2$ and initial velocity zero. After 5 s its fuel is finished and it decelerates with g. At the highest point its velocity becomes zero. Then it accelerates downwards with acceleration g and returns to ground. Plot velocity-time and displacement-time graphs for the complete journey.

Answer 1

In the graphs,

$v_A=a{t}_{OA}=\left(4\right)\left(5\right)=20m/s$

$v_B=0=v_A-g{t}_{AB}$

$\therefore t_{AB}=\frac{v_A}{g}=\frac{20}{10}=2s$

$\therefore t_{OAB}=(5+2)s=7s$

Now, $\therefore s_{OAB}$ =area under v-t graph between 0 to 7 s

$=\frac{1}{2}\left(7\right)\left(20\right)=70m$

Now, $\left|s_{OAB}\right|=\left|s_{BC}\right|=\frac{1}{2}g{t}_{BC}^2$

$\therefore 70=\frac{1}{2}\left(10\right)t_{BC}^2$

$\therefore t_{BC}=\sqrt{14}=3.7s$

$\therefore t_{OABC}=7+3.7=10.7s$

Also, $s_{OA}$ = area under v-t graph between OA

$=\frac{1}{2}\left(5\right)\left(20\right)=50m$